SOLUTION: what is the integral of sinxsin4x using double integration by parts?

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Question 1191872: what is the integral of sinxsin4x using double integration by parts?
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Let the integral that we wish to integrate be I

I%22%22=%22%22int%28sin%28x%29%2Asin%284x%29dx%29

Which factor, if either, is harder to integrate?
I'd say sin(4x) is harder to integrate, so let it be u.

matrix%282%2C1%2C%0D%0Au=sin%284x%29%2C%0D%0Adu=4cos%284x%29dx%29

I%22%22=%22%22int%28sin%28x%29%2Asin%284x%29dx%29%22%22=%22%22u%2Av-int%28v%2Adu%29%22%22=%22%22
-sin%284x%29cos%28x%29-int%28%28-cos%28x%29%5E%22%22%29%284cos%284x%29dx%5E%22%22%29%29%22%22=%22%22
-sin%284x%29cos%28x%29%2B4int%28cos%28x%29cos%284x%29dx%29

So let's write and mark what we have so far, so we can come back to
it and substitute:
-----------------------------------------------------------
I%22%22=%22%22-sin%284x%29cos%28x%29%2B4int%28cos%28x%29%28cos%284x%29dx%29%29
-----------------------------------------------------------
We now have find int%28cos%28x%29cos%284x%29dx%29

But we have to be careful here, so we don't "undo" what we just did.
For this part, be sure to let u equal to the part that came from
what we let u equal before.

matrix%282%2C1%2C%0D%0Au=cos%284x%29%2C%0D%0Adu=-4sin%284x%29dx%29

int%28cos%28x%29%2Acos%284x%29dx%29%22%22=%22%22u%2Av-int%28v%2Adu%29%22%22=%22%22
cos%284x%29sin%28x%29-int%28%28sin%28x%29%5E%22%22%29%28-4sin%284x%29dx%5E%22%22%29%29%22%22=%22%22
cos%284x%29sin%28x%29%2B4int%28sin%28x%29sin%284x%29dx%29

Let's write what we have here.
int%28cos%28x%29%2Acos%284x%29dx%29%22%22=%22%22cos%284x%29sin%28x%29%2B4int%28sin%28x%29sin%284x%29dx%29

That integral on the right is just the original integral that we called I:

int%28cos%28x%29%2Acos%284x%29dx%29%22%22=%22%22cos%284x%29sin%28x%29%2B4I

So we go back to

I%22%22=%22%22-sin%284x%29cos%28x%29%2B4int%28cos%28x%29%28cos%284x%29dx%29%29

and substitute  cos(4x)sin(x)+4I for the last integral:

I%22%22=%22%22-sin%284x%29cos%28x%29%2B4%28cos%284x%29%5E%22%22sin%28x%29%2B4I%29

I%22%22=%22%22-sin%284x%29cos%28x%29%2B4cos%284x%29%5E%22%22sin%28x%29%2B16I

Now we solve for I

-15I%22%22=%22%22-sin%284x%29cos%28x%29%2B4cos%284x%29sin%28x%29

I%22%22=%22%22expr%281%2F15%29sin%284x%29cos%28x%29-expr%284%2F15%29cos%284x%29sin%28x%29

But don't forget to put +C on the end.  
[It isn't necessary to put +C in the integration steps.]

I%22%22=%22%22expr%281%2F15%29sin%284x%29cos%28x%29-expr%284%2F15%29cos%284x%29sin%28x%29%2BC 

Edwin